如何等待所有异步任务在Node.js中完成？

Question

我的程序需要运行许多异步任务.

所有异步任务完成后,必须运行一项任务.

如何让这个函数等待所有异步函数？

let urls=[....]

for(var i=0; i<urls.length; i++){
    doRequest(urls[i])
}
finish()



function doRequest(url){
    request({
      uri: url,
    }, function(error, response, body) {
        //do some tasks
    });
}

function finish(){
    console.log('finish')
}

Answer 1

承诺正是您所需要的.它们现在在JS中是原生的,不需要额外的库.

function finish () {
    console.log("finish");
}

function doRequest (uri) {
    // wrap code in promise
    return new Promise((resolve, reject) => {
        request(
            { uri },
            (error, response, body) => {
                if (error) {
                    // error case, similar to "throw"
                    reject(error);
                }
                else {
                    // success case, similar to "return"
                    resolve({ response, body });
                }
            }
        );
    });
}

let urls=[...];

// Basic way to do it:
Promise.all(urls.map(doRequest)).then(finish);

// async/await notation:
// you must be in an "async" environement to use "await"
async function wrapper () {
    await Promise.all(urls.map(doRequest));
    finish();
}
// async function return a promise transparently
wrapper();